A bullet weighing 15 g, flying horizontally at a speed of 670 m / s, pierces a board 3.5 cm thick.
| May 2, 2021 | education
A bullet weighing 15 g, flying horizontally at a speed of 670 m / s, pierces a board 3.5 cm thick. Determine the kinetic erengia of a bullet after piercing a board if the resistance force when it moves in a tree is 74 kN.
Data: m (mass of a horizontally flying bullet) = 15 g = 0.015 kg; V0 (initial flight speed) = 670 m / s; h (thickness of the punched board) = 3.5 cm = 0.035 m; Fcopr (resistance force) = 74 kN = 74 * 10 ^ 3 N.
To calculate the kinetic energy of the bullet after piercing the board, we use the formula: A = Fcopr * h = Ek0 – Ek1, whence Ek1 = Ek0 – Fcopr * h = m * V0 ^ 2/2 – Fcopr * h.
Calculation: Ek1 = 0.015 * 670 ^ 2/2 – 74 * 10 ^ 3 * 0.035 = 776.75 J.
Answer: The kinetic energy became equal to 776.75 J.
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